Functions Exercise 1a Solutions The famous mathematician ” Lejeune Dirichlet” defined a function. Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to …
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]]>The famous mathematician ” Lejeune Dirichlet” defined a function.
Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.
Chapter 1 Functions Exercise 1a Solutions for inter first year students, prepared by Mathematics expert of www.basicsinmaths.com
I.
then find the values of (i) f (3) (ii) f (0) (iii) f (– 1.5) (iv) f (2) + f (– 2) (v) f (– 5 )
Sol:
Given
Domain of f(x) is (– 3, ∞)
(i) f (3)
3 lies in the interval x > 1
⟹ f(x) = x + 2
f(3) = 3 + 2 = 5
∴ f (3) = 5
(ii) f (0)
0 lies in interval – 1 ≤ x ≤ 1
⟹f(x) = 2
∴ f (0) = 2
(iii) f (– 1.5)
– 1.5 lies in interval – 3 < x < – 1
⟹f(x) = x – 1
f (– 1.5) = – 1.5 – 1 = – 2.5
∴ f (– 1.5) = – 2.5
(iv) f (2) + f (– 2)
2 lies in the interval x > 1
⟹ f (x) = x + 2
f (3) = 2 + 2 = 4
f (2) = 4
– 2 lies in interval – 3 < x < – 1
⟹f(x) = x – 1
f (– 2) = – 2 – 1 = – 3
f (– 2) = – 2 – 1 = – 3
now f (2) + f (– 2) = 4 – 3 = 1
∴ f (2) + f (– 2) = 1
(v) f (– 5)
since domain of f(x) is (– 3, ∞)
f (– 5) is not defined
Sol:
Given f: R – {0} ⟶ R is defined by f(x) =
f (1/x) =
Now
f (x) + f (1/x) =
∴ f (x) + f (1/x) = 0
Sol:
Given f: R ⟶ R is defined by f(x) =
f (tan θ) =
= cos 2θ
Sol:
Given f: R – {±1} ⟶ R is defined by f(x) =
Sol:
Given A = {– 2, – 1, 0, 1, 2} and f: A ⟶ B is a surjection defined by f(x) = x^{2} + x + 1
f(– 2) = (–2)^{2} + (–2) + 1
= 4 – 2 + 1 = 3
f(– 1) = (–1)^{2} + (–1) + 1
= 1 – 1 + 1 = 1
f(0) = (0)^{2} + (0) + 1
= 0 + 0 + 1 = 1
f(1) = (1)^{2} + (1) + 1
=1 +1 + 1 = 3
f( 2) = (2)^{2} + (2) + 1
= 4 + 2 + 1 = 4
∴ B = {1, 3, 7}
Sol:
Given A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) =
Sol:
Given f (x + y) = f (xy) ∀ x, y ∈ R
Let x = 0 and y = 0
f (0 + 0) = f (0 × 0) = f (0)
f (1) = f (0 + 1)
= f (0 × 1)
= f (0)
f (2) = f (1 + 1)
= f (1 × 1)
= f (1)
= f (0)
f (3) = f (1 + 2)
= f (1 × 2)
= f (2)
= f(0)
Similarly, f(4) = 0
f(5) = 0
and so on.
∴ f is a constant function
II.
(i) f: A ⟶ A (ii) g: A ⟶ A
Sol:
A = {– 1, 0, 1}; f: A ⟶ A
f (x) = x^{2}
f (– 1) = (– 1)^{2} = 1
f (0) = (0)^{2} = 0
f (1) = (1)^{2} = 1
∵ range is not equal to co domain of f
f is nor a surjection
A = {– 1, 0, 1}; g: A ⟶ A
g (x) = x^{3}
g (– 1) = (– 1)^{3} = – 1
g (0) = (0)^{3} = 0
g (1) = (1)^{3} = 1
∵ range is equal to co domain of f
f is a surjection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
2x_{1} + 1 = 2x_{2} + 1
2x_{1} = 2x_{2}
x_{1} = x_{2}
x_{1}, x_{2} ∈ R, f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y =
3y = 2x + 1
3y – 1 = 2x
⟹ x = ∈ R
Now
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
x_{1} = x_{2}
x_{1}, x_{2} ∈ R, f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y = 2^{x}
x = ∈ (0, ∞)
Now f(x) = 2^{x}
=
= y
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ (0, ∞)
f(x_{1}) = f(x_{2})
x_{1} = x_{2}
x_{1}, x_{2} ∈ (0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y =
x = e^{y} ∈ (0, ∞)
Now f(x) =
=
= y
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ [0, ∞)
f(x_{1}) = f(x_{2})
x_{1} = x_{2} [∵ x_{1}, x_{2} ∈ [0, ∞)]
x_{1}, x_{2} ∈ [0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y = x^{2}
x = ∈ [0, ∞)
Now f(x) = x^{2}
=
= y
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
x_{1} = ± x_{2} [∵ x_{1}, x_{2} ∈ R]
x_{1}, x_{2} ∈ [0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} ≠ x_{2}
∴ f is not an injection
Let y = f(x)
y = x^{2}
x = ∈ R
Now f(x) = x^{2}
=
= y
∴ f is surjection
f is not an injection but surjection
∴ f is not a bijection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
x_{1} = ± x_{2} [∵ x_{1}, x_{2} ∈ R]
x_{1}, x_{2} ∈ [0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} ≠ x_{2}
∴ f is not an injection
f (1) = 1^{2} = 1
f (– 1) = (–1)^{2} = 1
here ‘– 1’ has no pre image
∴ f is not a surjection
f is not an injection and not surjection
∴ f is not a bijection
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]]>TS Polycet || Solved Previous Question Papers 2020 Math The State Board of Technical Education and Training (SBTET), Telangana, Hyderabad will conduct “Polytechnic Common Entrance Test (POLYCET)” for the candidates seeking admission in to all Diploma Courses in Engineering /Non Engineering Technology. The Main Subjects in this Exam are Maths, Physics, Chemistry and Biology. Here …
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1.If 7 divides a^{2} then
a^{2} ను 7 భాగించినచో
(1) 7 divides a (a ను 7 భాగిస్తుంది)
(2) 7 divides ( ను 7 భాగిస్తుంది)
(3) a divide 7 (7 ను a భాగిస్తుంది)
(4) none (ఏదీ కాదు)
Answer: (1)
2. In the formula , which of the following is true?
అయిన, ఈ క్రింది వాటిలో ఏది సత్యము.
(1) x > 0, y > 0, a = 1
(2) x < 0, y < 0, a = 1
(3) a > 0, y > 0, x = 1
(4) x > 0, y > 0, a ≠ 1
Answer: (4)
Answer: (3)
Answer: (4)
Answer: (2)
TS 10th class maths concept (E/M)
TS 10th Class Maths Concept (T/M)
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]]>TS Polycet || Solved Previous Question Papers 2021 Maths The State Board of Technical Education and Training (SBTET), Telangana, Hyderabad will conduct “Polytechnic Common Entrance Test (POLYCET)” for the candidates seeking admission in to all Diploma Courses in Engineering /Non Engineering Technology. The Main Subjects in this Exam are Mathematics, Physics, Chemistry and Biology. Here …
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]]>
Answer: (2)
Answer: (4)
(1) Natural numbers (సహజ సంఖ్య)
(2) Irrational number (కరణీయ సంఖ్య)
(3) Rational Number (అకరణీయ సంఖ్య)
(4) An Integer (పూర్ణ సంఖ్య)
Answer: (2)
Answer: (3)
Answer: (2)
<< Page 1 of 14 >>
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]]>Power point presentations About a Presentation? A communication device that relays a topic to an audience in the form of a slide show, a demonstration, a lecture or speech where words and pictures intend to complement each other. Different Types of Power Point Presentations: 1) Informative Presentations 2) Instructive Presentations 3) Persuasive Presentations 4) …
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]]>Power point presentations
A communication device that relays a topic to an audience in the form of a slide show, a demonstration, a lecture or speech where words and pictures intend to complement each other.
1) Informative Presentations
2) Instructive Presentations
3) Persuasive Presentations
4) Motivational Presentations
5) Decision-making Presentations
6) Progress Presentations
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]]>TS Tet 2022 syllabus For Mathametic Paper 2 is very useful for writing candidates for TS TET. Knowing the syllabus will make the practice of the candidates easier. TS Tet 2022 Syllabus For Mathametic Paper 2 TS TET రాసె అభ్యర్థులు చాలా ఉపయోగపడుతుంది. సిలబస్ తెలియడం వల్ల అభ్యర్థుల అభ్యసన సులువు అవుతుంది. free Syllabus Telangana Teachers …
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1 . ప్రధాన సంఖ్యలు మరియు సంయుక్త సంఖ్యలు (Prime and Composite numbers)
2. భాజనీయత సూత్రాలు (Test of divisibility Rules)
3. సంఖ్యల రకాలు (types of numbers)
4. వాస్తవ సంఖ్యలు (Real numbers)
5. భిన్నాలు మరియు దశాంశ భిన్నాలు (Fractions and Decimal Fractions)
6. క .సా .గు . మరియు గ . సా .భా. యూక్లిడ్ భాగహార పద్ధతి (LCM and HCF – Euclid division Lemma)
7. వర్గాలు – వర్గ మూలాలు (Squares – Square roots)
8. ఘనాలు – ఘన మూలాలు (Cubs – Cube roots
9. సంఖ్యల అమరిక మరియు సంఖ్యల ఫజిల్ (Numbers Pattern and Numbers puzzle)
10. యూక్లిడ్ భాజనీయత సూత్రం (Euclid Division lemma)
11. సంవర్గమానము (Logarithms)
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]]>TS Intermediate First Year Maths Question Papers 2022 PDF As per reduced syllabus were designed by the ‘Basics in Maths‘ team. These Model papers 2022 to do help the intermediate First-year Maths students. TS Inter Maths 1A and 1B Modelpapers 2022 as per reduced syllabus are very useful in IPE examinations. Ts Inter …
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]]>As per reduced syllabus were designed by the ‘Basics in Maths‘ team.
These Model papers 2022 to do help the intermediate First-year Maths students.
TS Inter Maths 1A and 1B Modelpapers 2022 as per reduced syllabus are very useful in IPE examinations.
TS Inter Maths 1A Model papers 2022 as per reduced syllabus were designed by the ‘Basics in Maths‘ team.
These Model papers 2022 to do help the intermediate First-year Maths students.
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]]>10 Maths ICSE Banking MCQS With Answers TS 10th class maths concept (E/M) 1. Mr Gupta gets ₹ 6,455 at the end of the one year at the rate of 14% per annum in a recurring deposit account find the monthly instalment A) ₹ 200 B) ₹ 600 C) ₹ 500 D) ₹ …
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A) ₹ 200
B) ₹ 600
C) ₹ 500
D) ₹ 300
A) ₹ 7865
B) ₹ 12875
C) ₹875
D) ₹13865
A) ₹ 78650
B) ₹ 2350
C) ₹12350
D) ₹13050
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]]>Real Numbers 1. Insert 4 rational numbers between 1 and without using formula Sol: we have to insert 4 rational numbers between 1 and and 1 4 +1 = 5 1 can be written as Lcm of (4, 5) = 20 2. The prime factorisation of a natural number (n) is 23 × 32 …
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]]>Sol: we have to insert 4 rational numbers between 1 and and 1
4 +1 = 5
1 can be written as
Lcm of (4, 5) = 20
2. The prime factorisation of a natural number (n) is 2^{3} × 3^{2} × 5^{2} × 7. How many consecutive zeros will it have at the end of it? Justify your answer.
Sol:
n = 2^{3} × 3^{2} × 5^{2} × 7
= 2 × 2^{2} × 3^{2} × 5^{2} × 7
= 2 × 3^{2} × 7 (5^{2} × 2^{2})
= 2 × 3^{2} × 7 × (10)^{2} = 12600
∴ n has 2 consecutive zeros at the end
3. Find the value of
Sol:
4. Write any two irrational numbers between 3 and 4
Sol:
5. Find the value of
6. Find the HCF and LCM of 90, 144 by using the prime factorisation method
Sol:
7. Is rational or irrational? justify your answer
Sol:
8. Expand
Sol:
9. Find the HCF of 24 and 33 by using division method
Sol:
33 > 24
33 = 24 × 1 + 9
24 = 9 × 2 + 6
9 = 6 × 1 + 3
6 = 3 × 2 + 0
∴ HCF of 24 and 33 = 3
10. Find the value of
Sol:
11. Ramu says, “if = 0, the value of x is 0”. Do you agree with him? Give reason.
Sol:
Sol: let 10, 12, 16 be three two-digit numbers
Prime factarisation of 10 = 2 × 5 = 2^{1}× 5^{1}
Prime factarisation of 12 = 2 × 2× 3 = 2^{2} × 3
Prime factarisation of 16 = 2 × 2× 2× 2 = 2^{4}
HCF of 10, 12 and 16 = 2^{1} = 2
LCM of 10, 12 and 16 = 2^{4} × 3× 5 = 8 × 3× 5 = 120
2. Give an example for each of the following
(i) The product of two irrational numbers is a rational number
(ii) The product of two irrational numbers is an irrational number
Sol:
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]]>TS 10th Chapter 4 Pair of Linear Equations in Two Variables, Parallel, coincident and intersecting lines. Dependent and independents lines An equation of the form ax + by + c = 0 where a, b, c are real numbers and a2 + b2 ≠ 0 is called a linear equation in two variables x and …
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]]>An equation of the form ax + by + c = 0 where a, b, c are real numbers and a^{2} + b^{2} ≠ 0 is called a linear equation in two variables x and y.
Two linear equations in the same variables are called a pair of linear equations in two variables
a_{1}x + b_{1}y + c_{1} = 0; a_{2}x + b_{2}y + c_{2} = 0 are the pair of linear equations in two variables in x and y.
For linear equations in two variables, there are infinitely many solutions.
Ex: x + y = 10
x =7, y = 3; x = 6, y = 4; x = 1, y = 9; x =2, y = 8; x=3, y = 7 like we have infinitely many solutions.
⇰ For finding exact values of x and y we have to know two linear equations.
⇰ A pair linear equations in two variables solved by four methods
Solving the pair of linear equations in two variables by using graphical Method:
After plotting the points in the above tables in Cartesian plane, we observe that two straight lines intersect at the point (2, 1)
There is only one solution for this pair of linear equations in two variables.
These equations are known as consistent pair of linear equations and they have a unique solution.
After plotting the points in the above tables in Cartesian plane, we observe that two straight lines are parallel
There is no solution for this pair of linear equations in two variables.
These equations are known as inconsistent pair of linear equations and they have no solution.
After plotting the points in the above tables in Cartesian plane, we observe that two straight lines are coincide
There are infinitely many solutions for this pair of linear equations in two variables.
These equations are known as consistent pair of linear equations and they have infinitely solution.
Consistent and inconsistent:
If the system of equations has a solution, then they are consistent. If the system of equations has no solution, then they are inconsistent.
The relationship between coefficients and the nature of the equation system:
1. Draw the graph of the following pair of linear equations in two variables and find their solution from the graph
3x – 2y = 2 and 2x + y = 6
The two lines intersect at the point (2, 2)
∴ solutions is x = 2, y = 2
2. Draw the graph of the following pair of linear equations in two variables and find their solution from the graph
x – 2y = –1 and 2x – y – 4 = 0
The two lines intersect at the point (3, 2)
∴ solutions is x = 3, y = 2
3. Represent the solution of linear equation graphically
x – 2y = –3 and 2x + y = 4
The two lines intersect at the point (1, 2)
∴ solutions is x = 1, y = 2
1. Neha went to a sale to purchase some points and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts is two less than twice the number of points purchased. Also, the number of number of skirts is four less than four times the number of pants purchased”.
Help her friend to find how many pants and skirts Neha bought.
Sol:
Let the number of points = x
and the number of skirts = y
the number of skirts is two less than twice the number points purchased
⟹ y = 2x – 2
the number of number of skirts is four less than four times the number of pants purchased
⟹ y = 4x – 4
The two lines intersect at the point (1, 0)
∴ solution is x = 1, y = 2
No. of Points = 1 and no. of skirts = 0
2. 10 students of a class X took part in a Mathematics quiz. If the no. of girls is 4 more than the no. of boys, then find the no, of boys and no. of girls who took part in the quiz.
Sol:
let the no. of boys = x
No. of girls = y
Total no. of students = 10
⟹ x + y = 10
The no. of girls is 4 more than the no. of boys
⟹ y = x + 4
⟹ x – y = – 4
The two lines intersect at the point (3, 7)
∴ solution is x = 3, y = 7
No. of boys = 3 and no. of girls = 7
3. Half the perimeter of a rectangular garden, whose length is 4m more than its width is 36 m. Find the dimensions of the garden.
Sol:
let the width of garden = x m.
Length of garden = y m.
Length of garden is 4m mote than its width
⟹ y = x + 4
x – y = – 4
half of the perimeter of rectangular garden is 36 m.
⟹ x + y = 36
⟹ x – y = – 4
The two lines intersect at the point (16, 20)
∴ solution is x = 16, y = 20
Width of garden = 16 m and Length of garden = 20 m
4. The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will be increased by 50 sq units. Find the length and breadth of rectangle.
Sol:
Let the length of rectangle = x units
the breadth of rectangle = y units
The area of rectangle = xy sq units
If the Length of rectangle is reduced by 5 units and breadth is increased by 2 units, then area will be reduced by 80 sq units.
⟹ (x – 5) (y + 2) = xy – 80
⟹ xy + 2x – 5y – 10 = xy – 80
⟹ 2x – 5y = – 70
If length is increased by 10 and breadth decreased by 5 then area will be increased by 50 squnits
⟹ (x + 10) (y – 5) = xy + 50
⟹ xy – 5x + 10y – 50 = xy + 50
⟹ – 5x + 10y = 100
The two lines intersect at the point (40, 30)
∴ solution is x = 40, y = 30
Length of rectangle = 40 units and Breadth of rectangle = 30 units
5. In X class, if three students sit on each bench one student will be left, if four students sit on each bench, one bench will be left. Find the number of students and number of benches in that class.
Sol:
Let the no. of students = x
The no. of benches = y
If three students sit on each bench one student will be left
⟹ 3y = x – 1
⟹ x – 3y = 1
If four students sit on each bench one bench will be left
⟹ 4 (y – 1) = x ⟹ 4y – 4 = x
⟹ x – 4y = – 4
The two lines intersect at the point (16, 5)
∴ solution is x = 16, y = 5
No. of students = 16 and No. of benches = 5
The relationship between coefficients and the nature of the equation system Related Problems:
1. Check whether the following pair of linear equations have a unique solution, infinitely many solutions or no solution
(i) 2x + 3y = 1; 3x – y = 7
(ii) 2x + 3y = 5; 4x + 6y = 10
(iii) 5x + 3y = 4;10x + 6y = 12
Sol:
(i) Given equations are x + 3y = 1; 3x – y = 7
a_{1} = 1, b_{1} = 3, c_{1} = 1; a_{2} = 3, b_{2} = –1, c_{2} = 7
∴ Given equations have unique solution
(ii) Given equations are 2x + 3y = 5; 4x + 6y = 10
a_{1} =2, b_{1} = 3, c_{1} = 5; a_{2} = 4, b_{2} = 6, c_{2} = 10
∴ Given equations have infinite solutions
(iii) Given equations are 5x + 3y = 4;10x + 6y = 12
a_{1} =5, b_{1} = 3, c_{1} = 4; a_{2} = 10, b_{2} = 6, c_{2} = 12
∴ Given equations have no solution
2. For what value of ‘p’ the following pair of equations has a unique solution
2x + py = – 5 and 3x + 3y = – 6
Sol:
Given equations are 2x + py = – 5, 3x + 3y = – 6
a_{1} =2, b_{1} = p, c_{1} = –5; a_{2} = 3, b_{2} = 3, c_{2} = –6
Above equations have a unique solution
∴ Given pair of linear equations have a unique solution when p ≠ 2
3. For what value of ‘p’ the following pair of equations has a unique solution
px + 3y –(p – 3)= 0 and 12x + py – p = 0
Sol:
Given equations are px + 3y – (p – 3) = 0, 12x + py – p = 0
a_{1} =p, b_{1} = 3, c_{1} = – (p – 3),; a_{2} =12, b_{2} = p, c_{2} = –p
Above equations have infinitely many solution
∴ Given pair of linear equations have infinitely many solutions when p = ±6
4. For what value of ‘k’ the following pair of equations represent coincident lines.
3x + 4y + 2 = 0 and 9x + 12y + k = 0
Sol:
Given equations are 3x + 4y + 2 = 0 and 9x + 12y + k = 0
a_{1} =3, b_{1} = 4, c_{1} = 2; a_{2} = 9, b_{2} = 12, c_{2} = k
Above equations are coincident lines
∴ Given pair of linear equations are coincident when k = 6
5. For what value of ‘k’ the following pair of equations represents the parallel lines
2x – ky + 3 = 0 and 4x + 6y – 5 = 0
Sol:
Given equations are 2x – ky + 3 = 0, 4x + 6y – 5 = 0
a_{1} =2, b_{1} = –k, c_{1} = 3; a_{2} = 4, b_{2} = 6, c_{2} = –5
Above equations represents parallel lines
∴ Given pair of linear equations represents parallel lines when k = – 3
In this method, we make one of the variables x or y as the subject from the 1^{st} equation (or 2^{nd} equation). Substitute this equation in 2^{nd} equation (or 1^{st} equation) and get the value of the variable involved, then by substituting this value in any one of the equations we get the value of second variable.
1. Solve the following pair of equations by using substitution method
2x – y = 5 and 3x + 2y = 11
Sol:
Given equations are 2x – y = 5 ———— (1)
3x + 2y = 11———– (2)
From (1)
y = 2x – 5
substitute y value in (2)
⟹ 3x + 2(2x – 5) = 11
3x + 4x – 10 = 11
7x = 11 + 10 = 21
x = 3
now y = 2x – 5
⟹ y = 2(3) – 5 = 6 – 5 = 1
∴ the solution is x = 3 and y = 1
2. Solve the following pair of equations by using substitution method
2x + 3y = 9 and 3x + 4y = 5
Sol:
Given equations are 2x + 3y = 9 ———— (1)
3x + 4y = 5———– (2)
From (1)
3y = 9 – 2x
∴ the solution is x = – 21 and y = 17
3. Solve the following pair of equations by using substitution method
3x – 5y = – 1 and x – y = – 1
Sol:
Given equations are 3x – 5y = – 1 ———— (1)
x – y = – 1———– (2)
From (2)
y = x + 1
substitute y value in (1)
⟹ 3x – 5(x + 1) = – 1
3x – 5x – 5= – 1
– 2x – 5= – 1
– 2x = – 1 + 5
– 2x = 4 ⟹ x = – 2
now y = x + 1
⟹ y = – 2 + 1 = – 1
∴ the solution is x = – 2 and y = – 1
4. Solve the following pair of equations by using substitution method
x + = 6 and 3x – = 5
Sol:
5. Solve the following pair of equations by using substitution method
0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3
Sol:
In this method first we eliminate one of the two variables by equating its coefficients. This give a single equation which can be solved to get the value of the other variable
Steps for Elimination Method:
1. Solve the following pair of equations by using Elimination method
2x – y = 5 and 3x + 2y = 11
2. Solve the following pair of equations by using Elimination method
8x + 5y = 9 and 3x + 2y = 4
3. Solve the following pair of equations by using Elimination method
3x + 4y = 25 and 5x – 6y = –9
4. Solve the following pair of equations by using Elimination method
2x + y = 5 and 3x – 2y = 4
5. Solve the following pair of equations by using Elimination method
3x + 2y = 11 and 2x + 4y =4
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